题意

给出 $n$ 个集合 $S_1 \cdots S_n$，$S_i = \{a_i\}$，每次给出 $x,y$，将第 $x$ 和第 $y$ 个元素所在的集合的最大值 $\div 2$，合并两个集合，然后输出新集合的最大值。

sol

每次求出两个集合，记录两个集合的最大值并删除，将两个集合与两个最大值除以 $2$ 后合并即可。考虑左偏树（可并堆），思路参见 [luoguP3377] 左偏树/可并堆
注意多测要清空

代码

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 100005;

struct Node {
    int l, r;
    int val;
    int dist;
}tr[N];

int n, m;
int fa[N];

int find(int x){
    if (x == fa[x]) return x;
    return fa[x] = find(fa[x]);
}

int merge(int x, int y){
    if (!x || !y) return x | y;
    if (tr[x].val < tr[y].val) swap(x, y);
    tr[x].r = merge(tr[x].r, y);
    if (tr[tr[x].l].dist < tr[tr[x].r].dist) swap(tr[x].l, tr[x].r);
    tr[x].dist = tr[tr[x].r].dist + 1;
    return x;
}

int main(){
    while (scanf("%d", &n) != EOF){
        for (int i = 1; i <= n; i ++ ) scanf("%d", &tr[i].val), tr[i].l = tr[i].r = tr[i].dist = 0, fa[i] = i;

        scanf("%d", &m);
        while (m -- ){
            int x, y;
            scanf("%d%d", &x, &y);

            int fx = find(x), fy = find(y);
            if (fx == fy) puts("-1");
            else {
                int rtx = merge(tr[fx].l, tr[fx].r), rty = merge(tr[fy].l, tr[fy].r);
                tr[fx].val /= 2, tr[fy].val /= 2;
                tr[fx].dist = tr[fy].dist = 0;
                int xl = tr[fx].l, xr = tr[fx].r, yl = tr[fy].l, yr = tr[fy].r;
                tr[fx].l = tr[fx].r = tr[fy].l = tr[fy].r = 0;
                fa[fx] = fa[fy] = fa[xl] = fa[xr] = fa[yl] = fa[yr] = merge(merge(rtx, fx), merge(rty, fy));
                printf("%d\n", tr[fa[fx]].val);
            }
        }
    }
}